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Ex 3.2, 12 - Chapter 3 Class 12 Matrices
Last updated at June 7, 2023 by Teachoo
Ex 3.2
Ex 3.2, 1
Ex 3.2, 2 (i)
Ex 3.2, 2 (ii) Important
Ex 3.2, 2 (iii)
Ex 3.2, 2 (iv)
Ex 3.2, 3 (i)
Ex 3.2, 3 (ii) Important
Ex 3.2, 3 (iii)
Ex 3.2, 3 (iv) Important
Ex 3.2, 3 (v)
Ex 3.2, 3 (vi) Important
Ex 3.2, 4
Ex 3.2, 5
Ex 3.2, 6
Ex 3.2, 7 (i)
Ex 3.2, 7 (ii) Important
Ex 3.2, 8
Ex 3.2, 9
Ex 3.2, 10
Ex 3.2, 11
Ex 3.2, 12 Important You are here
Ex 3.2, 13 Important
Ex 3.2, 14
Ex 3.2, 15
Ex 3.2, 16 Important
Ex 3.2, 17 Important
Ex 3.2, 18
Ex 3.2, 19 Important
Ex 3.2, 20 Important
Ex 3.2, 21 (MCQ) Important
Ex 3.2, 22 (MCQ) Important
Chapter 3 Class 12 Matrices
Serial order wise
Transcript
Ex 3.2, 12 Given 3[■8(x&y@z&w)] = [■8(x&6@−1&2w)] + [■8(4&x+y@z+w&3)] find the values of x, y, z and w. 3[■8(x&y@z&w)] = [■8(x&6@−1&2w)] + [■8(4&x+y@z+w&3)] [■8(3x&3y@3z&3w)] = [■8(x+4&6+x+y@1−z+w&2w+3)] Since matrices are equal. Corresponding elements are equal 3x = x + 4 3y = 6 + x + y 3z = 1 – z + w 3w = 2w + 3 Solving equation (1) 3x = x + 4 3x – x = 4 2x = 4 x = 4/2 x = 2 Solving equation (2) 3y = 6 + x + y 3y – y = 6 + x 2y = 6 + x Putting x = 2 2y = 6 + 2 2y = 8 y = 8/2 y = 4 Solving equation (4) 3w = 2w + 3 3w – 2w = 3 w = 3 Solving equation (3) 3z = – 1 + z + w 3z – z = – 1 + w 2z = – 1 + w Putting w = 3 2z = – 1 + 3 2z = 2 z = 2/2 z = 1 Hence, x = 2, y = 4 , w = 3 & z = 1
